Modellen & SimulatieSlides bij Hoorcollege 3

Programma

• Herhaling stabiliteit van scalaire DVs (incl. Verhulst)

• Slinger model

• Faseruimte

• Lineaire DVs in twee dimensies:

• Eigenwaarde decompositie

• Eigenwaardeprobleem

• Categorizatie van evenwichten

4 CHAPTER 1. MODELLING

If n(0) = n0, then c = n0/(1� n0) and the solution can be written

n(t) =n0e

rt

(1� n0) + n0ert

.

If r > 0 it can be checked thatlim

t!1n(t) = 1

for any initial state n0 > 0. That is, the population eventually saturates (N ! N

⇤) for any initialpopulation. This behavior is shown in the left plot of Figure 1.2.

Next, suppose that in attempt to control the aphid population, a predator is introduced intothe rose farm. One nasty predator is the ladybug (Fig. 1.1b. If you are an aphid, the ladybug isanything but a symbol of nonviolence!). Normally lady bugs will not bother eating aphids unlessthere is su�cient supply. The logistic model can be extended to incorporate predation by ladybugs as follows:

dn

dt

= rn

⇣1� n

s

⌘� n

2

1 + n

2. (1.5)

The last term gradually “turns on” ladybug predation as the population n grows greater than 1.We have re-introduced the saturation level s because there are now two important populationslevels—the natural saturation level, and the level for which ladybugs get “interested”. Somesolutions of this model are shown in the right plot of Figure 1.2. It is interesting to note that fors = 20 and either r = 0.1 or r = 0.6, the aphid population eventually tends to a constant value,independent of the initial population. This is similar to the case of the logistic equation withoutpredation (1.4). For the case s = 20, r = 0.25, however, the final aphid population depends onthe starting population. For a starting population of about n0 < 5, the aphids settle down toa low level of about n = 0.5, whereas for a starting population of n0 > 5, the aphid populationeventually stabilizes at a relatively high level of around n = 14.5. Such a change in the quality ofsolutions with a change in parameter values is called bifurcation.

0 5 10 15 20 25 30 35 400

0.5

1

1.5

t

n

Logistic model, r = 0.25

0 10 20 30 40 50 600

2

4

6

8

10

12

14

16

18

20

t

n

Logistic model with predation, s=20, r = {0.1, 0.25, 0.6}

Figure 1.2: Solutions of the logistic model. On the left, solutions of (1.4) with r = 0.25 for di↵erentinitial conditions. On the right, with predation (1.5), s = 20 and r = 0.6 (green), r = 0.25 (black),and r = 0.1 (red). For r = 0.25, there are two steady states, depending on the initial condition

This model is still rather crude, because it assumes an endless supply of ladybugs. An alterna-tive model might consider an isolated system of aphids and ladybugs, and model the populationsof both. Let p(t) denote the (normalized) population of ladybugs. In the absence of aphids, theladybugs would die out exponentially: dp/dt = �mp for some m > 0. But with an increasingnumber of aphids, the ladybugs will flourish, so the ladybug population should satisfy

dp

dt

= anp�mp

Verhulst model, r=1/4, Pc=1

P(t)

t

Programma

• Herhaling stabiliteit van scalaire DVs (incl. Verhulst)

• Slinger model

• Faseruimte

• Lineaire DVs in twee dimensies:

• Eigenwaarde decompositie

• Eigenwaardeprobleem

• Categorizatie van evenwichten

Vector Field (✓̇, v̇)✓(t) v(t)

✓̇ = v

v̇ = � sin ✓Pendulum

Phase plot✓(t) v(t) (✓(t), v(t))

✓̇ = v

v̇ = � sin ✓Pendulum

Phase plot✓(t) v(t) (✓(t), v(t))

✓̇ = v

v̇ = � sin ✓Pendulum

Phase plot✓(t) v(t) (✓(t), v(t))

✓̇ = v

v̇ = � sin ✓Pendulum

Phase plot (✓(t), v(t))

✓̇ = v

v̇ = � sin ✓Pendulum

d = det

s = spoor

d = s2/4

�1,2 2 R

zadelpunt

knoop

spiraal�1 = �̄2 2 C

cent

rum

punt

Phase portraitVector Field

�1 = �1.5, �2 = �0.8 v1 =

✓10

◆, v2 =

✓01

◆

Stable node

Phase portraitVector Field

v1 =

✓10

◆, v2 =

✓01

◆

Unstable node

�1 = 1.5, �2 = 0.8

v1 =

✓10

◆, v2 =

✓11

◆

Phase portraitVector Field

�1 = �1.5, �2 = �0.8

Stable node

d = det

s = spoor

d = s2/4

�1,2 2 R

zadelpunt

knoop

spiraal�1 = �̄2 2 C

cent

rum

punt

Phase portraitVector Field

v1 =

✓10

◆, v2 =

✓01

◆�1 = �1.5, �2 = 0.8

Saddle point

v1 =

✓10

◆, v2 =

✓11

◆

Phase portraitVector Field

�1 = �1.5, �2 = 0.8

Saddle point

d = det

s = spoor

d = s2/4

�1,2 2 R

zadelpunt

knoop

spiraal�1 = �̄2 2 C

cent

rum

punt

Phase portraitVector Field

�1,2 = �0.5± 1.0 i

Stable spiral (focus)

Phase portraitVector Field

�1,2 = �0.5± 1.0 i

Stable spiral (focus)

d = det

s = spoor

d = s2/4

�1,2 2 R

zadelpunt

knoop

spiraal�1 = �̄2 2 C

cent

rum

punt

Phase portraitVector Field

�1,2 = ±2i

Center point

Phase portraitVector Field

�1,2 = ±2i

Center point

d = det

s = spoor

d = s2/4

�1,2 2 R

zadelpunt

knoop

spiraal�1 = �̄2 2 C

cent

rum

punt

Phase portraitVector Field

v1 =

✓10

◆, v2 =

✓01

◆�1 = �1, �2 = 0

Singular case

Phase portraitVector Field

�1 = �1, �2 = 0

Singular matrix A =

�1 01 0

�

Phase portraitVector Field

v1 =

✓10

◆, v2 =

✓01

◆�1 = �2 = �1

Stable singular node A =

�1 00 �1

�

Phase portraitVector Field

�1 = �2 = �1 v1 =

✓01

◆

Stable degenerate node A =

�1 10 �1

�

Phase portraitVector Field

�1 = �2 = �1

Stable degenerate node